java - 在Android/java中求解两个方程组

Question

我有两个方程(直线公式的距离和斜率)

d = sqrt( (x2 - x1)^2 + (y2 - y1)^2 )
m = (y2 - y1)/(x2 - x1)

已知:d、m、x1、y1
未知:x2，y2

问题是距离方程不是线性的......

有没有办法用java编码(使用Android兼容库)来解决这个问题？我尝试做简单的猜测，但速度太慢。

谢谢

编辑:三角形代码

        Point p1 = new Point();
        Point p2 = new Point();
        projection.toPixels(gp1, p1);
        projection.toPixels(gp2, p2);

        Point p3 = new Point();
        double slope = (p2.y - p1.y) / (p2.x - p1.x);
        double x = 0;
        if (p2.y - p1.y >= 0 && p2.x - p1.x >= 0) {
            x = - Math.sqrt(600 / (1 + slope*slope)) + p2.x;
        } else if (p2.y - p1.y >= 0 && p2.x - p1.x < 0) {
            x = Math.sqrt(600 / (1 + slope*slope)) + p2.x;
        } else if (p2.y - p1.y < 0 && p2.x - p1.x >= 0) {
            x = - Math.sqrt(600 / (1 + slope*slope)) + p2.x;
        } else if (p2.y - p1.y < 0 && p2.x - p1.x < 0) {
            x = Math.sqrt(600 / (1 + slope*slope)) + p2.x;
        }
        double y = -slope*p2.x + slope*x + p2.y;

        p3.set((int) x, (int) y);

        double inverseSlope = 0;
        if (slope == 0) {
            inverseSlope = Double.MAX_VALUE;
        } else {
            inverseSlope = -1 / slope;
        }

        x = -Math.sqrt(300 / (1 + inverseSlope*inverseSlope)) + p3.x;
        y = -Math.sqrt(300 / (1 + inverseSlope*inverseSlope))*inverseSlope + p3.y;

        Point p4 = new Point();
        p4.set((int) x, (int) y);

        x = Math.sqrt(300 / (1 + inverseSlope*inverseSlope)) + p3.x;
        y = Math.sqrt(300 / (1 + inverseSlope*inverseSlope))*inverseSlope + p3.y;
        Point p5 = new Point();
        p5.set((int)x, (int) y);
        Path path = new Path();
        path.moveTo(p2.x, p2.y);
        path.lineTo(p4.x, p4.y);
        path.moveTo(p4.x, p4.y);
        path.lineTo(p5.x, p5.y);
        path.moveTo(p5.x, p5.y);
        path.lineTo(p2.x, p2.y);
        path.moveTo(p2.x, p2.y);
        canvas.drawPath(path, mPaint);

看来这是由斜率始终是整数引起的，因此当它 < 1 时，它是 0，这不好......

Answer 1

请复习下面的代数:

定义

x = x2-x1

和

y = y2-y1

然后

米 * x = y

和

d^2 = x^2 + m^2 * x^2 = (1 + m^2) * x^2

因此

x = sqrt(d^2/(1 + m^2))

然后

x2 - x1 = sqrt(d^2/(1 + m^2))

所以

x2 = sqrt(d^2/(1 + m^2) + x1

同样

y = sqrt(d^2/(1 + m^2)) * m

y2 = sqrt(d^2/(1 + m^2)) * m + y1

所以答案是:

x2 = sqrt(d^2/(1 + m^2)) + x1

y2 = sqrt(d^2/(1 + m^2)) * m + y1